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Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.

However, I just cannot get a proper proof.

Quanto
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djsg
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    Use the mean value theorem and the fact that the derivative of $r\to\sqrt{r}$ is decreasing. – GReyes Feb 20 '21 at 03:59
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    Here's one approach: it suffices to show that $f(x) = \sqrt{x} - \sqrt{x-1}$ is decreasing for $x \geq 1$. To do this, you can just prove that $f'(x)$ is negative for $x \geq 1$! – diracdeltafunk Feb 20 '21 at 03:59
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    $\sqrt{x+1} - \sqrt{x} = \frac{1}{\sqrt{x+1} + \sqrt{x}}$ decreases with increasing $x$, compare https://math.stackexchange.com/q/545704/42969 or https://math.stackexchange.com/q/1312776/42969. – Or use that $\sqrt x$ is a concave function. – Martin R Feb 20 '21 at 04:01

5 Answers5

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Note, for $x\ge 1$ $$x>\sqrt{x^2-1}$$ $$4x>2x+2\sqrt{x^2-1}=(\sqrt{x+1}+\sqrt{x-1})^2$$ $$2\sqrt x>\sqrt{x+1}+\sqrt{x-1} $$ $$\sqrt x -\sqrt{x-1}> \sqrt{x+1}-\sqrt x$$

Quanto
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Alternative approach that avoids Calculus:
square both sides.

LHS squared is $(2x-1) - 2\sqrt{x(x-1)}$
RHS squared is $(2x+1) - 2\sqrt{x(x+1)}$

So, taking LHS - RHS, the question reduces to whether
$-2 + 2\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 0.$

This is equivalent to asking whether
$\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 1.$

Edit
Stealing a caveat from Quanto's answer, note that when $x\geq 1$, the LHS above is clearly positive.

Again squaring both sides, the problem reduces to asking whether
$x[2x - 2\sqrt{x^2 -1}] > 1.$

This is answered by applying the scalar of $[2x + 2\sqrt{x^2 -1}],$ which is a positive scalar, to both sides, so that the problem is transformed to asking whether

$$x[2x - 2\sqrt{x^2 -1}][2x + 2\sqrt{x^2 -1}] > [2x + 2\sqrt{x^2 -1}].\tag1$$

Since the LHS of equation (1) above simplifies to $4x$,
and since it is clear that $2\sqrt{x^2 - 1} < 2x$
the problem is resolved.

user2661923
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\begin{align} \sqrt{x} - \sqrt{x-1} &= \dfrac{(\sqrt{x} - \sqrt{x-1})(\sqrt{x} + \sqrt{x-1})}{\sqrt{x} + \sqrt{x-1})} \\ &= \dfrac{x-(x-1)}{\sqrt{x} + \sqrt{x-1}} \\ &= \dfrac{1}{\sqrt{x} + \sqrt{x-1}} \end{align}

\begin{align} \sqrt{x+1} - \sqrt{x} &= \dfrac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x})} \\ &= \dfrac{(x+1)-x}{\sqrt{x+1} + \sqrt{x}} \\ &= \dfrac{1}{\sqrt{x+1} + \sqrt{x}} \end{align}

So \begin{align} \sqrt{x+1} &> \sqrt{x-1} \\ \sqrt{x+1} + \sqrt x &> \sqrt x + \sqrt{x-1} \\ \left(\dfrac{1}{\sqrt{x+1} + \sqrt x}\right) &< \left(\dfrac{1}{\sqrt x + \sqrt{x-1}}\right) \\ \sqrt{x+1} - \sqrt x &< \sqrt x - \sqrt{x-1} \\ \end{align}

Steven Alexis Gregory
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$$\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}\quad| +\sqrt{x}$$ $$2\sqrt{x} > \sqrt{x-1}+\sqrt{x+1} \quad|^2$$ $$4x>2x+ 2\sqrt{(x-1)(x+1)} \quad| -2x $$ $$2x>2\sqrt{(x-1)(x+1)} \quad| /2$$ $$x=\sqrt{x^2-1} \quad|^2$$ $$x^2>x^2-1 \quad|-x^2$$ $$0>-1$$ The transfoemations are valid in both directions, top down and bottom up

miracle173
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This is not a proof.

When $x$ is large $$\Big[\sqrt{x} - \sqrt{x-1}\Big]-\Big[\sqrt{x+1} - \sqrt{x}\Big]=\frac 1{4 x^{\frac 32}}\sum_{n=0}^\infty \frac {a_n}{x^{2n}}$$ all coefficients being positive make the decreasing sequence $$\left\{1,\frac{5}{16},\frac{21}{128},\frac{429}{4096},\frac{2431}{32768},\frac{29393 }{524288},\frac{185725}{4194304},\cdots\right\}$$

Claude Leibovici
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